Question: A particle moves in the $xy$ -plane so that at any time $t\geq 0$ its position vector is $(4t^3-6t^2,3t^2+2t)$. What is the particle's velocity vector at $t=2$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $(4,8)$ (Choice B) B $(24,14)$ (Choice C) C $(8,16)$ (Choice D) D $(16,10)$
Background When solving motion problems, it's important to remember the relationship between the position vector $(x,y)$, the velocity vector $\vec{v}(t)$, and the acceleration vector $\vec{a}(t)$ of the moving particle: $\vec{v}(t)=\left(\dfrac{dx}{dt},\dfrac{dy}{dt}\right)$ $\vec{a}(t)=\dfrac{d}{dt}\vec{v}(t)=\left(\dfrac{d^2x}{dt^2},\dfrac{d^2y}{dt^2}\right)$ Setting up the math We are given that the particle's position vector is $(4t^3-6t^2,3t^2+2t)$. We are asked to find the particle's velocity vector at $t=2$. In other words, we need to find $\vec{v}(2)$. Finding $\vec{v}(t)$ $\begin{aligned} \vec{v}(t)&=\left(\dfrac{d}{dt}(4t^3-6t^2),\dfrac{d}{dt}(3t^2+2t)\right) \\\\ &=(12t^2-12t,6t+2) \end{aligned}$ Finding $\vec{v}(2)$ $\begin{aligned} \vec{v}({2})&=(12({2})^2-12(2),6({2})+2) \\\\ &=(48-24,12+2) \\\\ &=(24,14) \end{aligned}$ In conclusion, the particle's velocity vector at $t=2$ is $(24,14)$.